91Ó°ÊÓ

Actual mass of ${}^1\mathrm{H}$is ${\mathrm{m}}_{\mathrm{H}}=1.007825\mathrm{u}$.

Actual mass of a neutron is ${\mathrm{m}}_{\mathrm{n}}=1.008665\mathrm{u}$.

Actual mass of ${}^{120}\mathrm{Sn}$is${\mathrm{m}}_{\mathrm{n}}=119.902197\mathrm{u}$ .

Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. It may sometimes be called mass excess. The mass excess of a nuclide is the difference between its actual mass and its mass number in atomic mass units. Now, using the calculated mass excess for each given nuclide, we can convert it into the required unit of MeV/c².

Formulae:

The mass excess of a nuclide,$\mathrm{Mass}\mathrm{excess}\left(\mathrm{D}\right)=\mathrm{Actual}\mathrm{mass}-\mathrm{Mass}\mathrm{in}\mathrm{amu}..........\left(1\right)$

The conversion of atomic mass unit into$\mathrm{MeV}/{\mathrm{c}}^2,1\mathrm{amu}=931.5\mathrm{Mev}/{\mathrm{c}}^2.................\left(2\right)$ ,

Using the given data in equation (1), the mass excess of in atomic mass units is given as follows:

$$\begin{gathered} {∆}_1=1.007825\mathrm{u}-1.000000\mathrm{u} \\ =0.007825\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is 0.007825 u.

Now, using the result of part (a) and equation (2), we can get the mass excess of in$\mathrm{MeV}/{\mathrm{c}}^2$ as follows:

$$\begin{gathered} {∆}_1=0.007825\mathrm{u}×\left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =+7.290\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is +7.290 MeV/${\mathrm{c}}^2$ .

Using the given data in equation (1), the mass excess${∆}_{\mathrm{n}}$of a neutron in atomic mass units is given as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=1.008665\mathrm{u}-1.000000\mathrm{u} \\ =0.008665\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is 0.008665 u.

Now, using the result of part (a) and equation (ii), we can get the mass excess${∆}_{\mathrm{n}}$ of a neutron in as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=0.008665\mathrm{u}×\left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =+8.071\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is $+8.071\mathrm{MeV}/{\mathrm{c}}^2$.

Using the given data in equation (1), the mass excess ${∆}_{\mathrm{Sn}}$of ${}^{120}\mathrm{Sn}$in atomic mass units is given as follows:

$$\begin{gathered} {∆}_{\mathrm{Sn}}=119.902197\mathrm{u}-120.000000\mathrm{u} \\ =-0.09780\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is -0.9780 u.

Now, using the result of part (a) and equation (2), we can get the mass excess${∆}_{\mathrm{Sn}}$ of${}^{120}\mathrm{Sn}$ in$\mathrm{MeV}/{\mathrm{c}}^2$ as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=-0.09780\mathrm{u}×\left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =-91.10\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is$-91.10\mathrm{MeV}/{\mathrm{c}}^2$ .